Problem: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x + 8)^2 - 1$ $\text{lesser }x = $
Explanation: $\begin{aligned} (x + 8)^2 - 1&= 0 \\\\ (x+8)^2&=1 \\\\ \sqrt{(x+8)^2}&=\sqrt{1} \end{aligned}$ $\begin{aligned} x+8&=\pm1 \\\\ x&=\pm1-8 \\ \phantom{(x + 8)^2 - 1}& \\ x=-9&\text{ or }x=-7 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -9 \\\\ \text{greater } x &= -7 \end{aligned}$